Letting $ x=\cos(\pi \alpha)$ and the equation turns into $ (2x+1)(2x^{2}+x-2)=0$. The only plausible solutions are $ x=\frac{-1}{2}, \frac{-1+\sqrt{17}}{4}$. $ x=-\frac{1}{2}\Rightarrow \alpha=\frac{2}{3}$ but how would you prove that $ \frac{\arccos(\frac{-1+\sqrt{17}}{4})}{\pi}$ is irrational?

Solution by olorin: Ok, I have a solution using cyclotomic fields. $ t: = \cos\pi x$, then $ \cos 2\pi x = 2t^{2} - 1$ and $ \cos 3\pi x = 4t^{3} - 3t$. So $ 0 = 4t^{3} + 4t^{2} - 3t - 2 = (2t + 1)(2t^{2} + t - 2)$ and $ t = - {1\over 2}$ or $ 2t^{2} + t - 2 = 0$. $ x\mapsto \cos\pi x$ is a bijection $ [0,1]\to [ - 1,1]$ with $ \cos{2\pi\over 3} = - {1\over 2}$. So $ \cos\pi x = t = - {1\over 2}$ and $ x\in[0,1]$ forces $ x = {2\over 3}$. Assume therefore $ 2t^{2} + t - 2 = 0$ and $ t = { - 1\pm\sqrt {17}\over 4}$ and $ [\mathbb{Q}(t): \mathbb{Q}] = 2$. Write $ {x\over 2} = {m\over n}$ with $ m\in\mathbb{Z},n\in\mathbb{N},\gcd (m,n) = 1$. Then $ \zeta: = e^{2m\pi i/n}$ is a primitive $ n$-th root of unity with $ \zeta + \zeta^{ - 1} = 2\cos{2m\pi \over n} = 2t$. So $ \zeta$ is root of $ (X - \zeta)(X - \zeta^{ - 1}) = X^{2} - 2tX + 1\in\mathbb{Q}(t)[X]$ and $ [\mathbb{Q}(\zeta): \mathbb{Q}(t)]\leq 2$, and so $ \varphi (n) = [\mathbb{Q}(\zeta): \mathbb{Q}] = [\mathbb{Q}(\zeta): \mathbb{Q}(t)][\mathbb{Q}(t): \mathbb{Q}]\leq 4$. If $ p|n$ for some prime $ p\geq 7$, then $ 4 < p - 1|\varphi (n)$. Contradiction! If $ p^{2}|n$ for some prime $ p\geq 3$, then $ 4 < (p - 1)p|\varphi (n)$. Contradiction! If $ pq|n$ for some primes $ p > q\geq 3$, then $ 4 < (p - 1)(q - 1)|\varphi (n)$. Contradiction! So $ n = 2^\alpha$ or $ n = 2^\alpha\cdot 3$ or $ n = 2^\alpha\cdot 5$ for some $ \alpha\in\mathbb{N}_{0}$. This gives $ n\in\{1,2,2^{2},2^{3},3,2\cdot 3,2^{2}\cdot 3,5,2\cdot 5\}$ and then $ t = \cos{2m\pi \over n}\in\{0,\pm 1,\pm{1\over 2},\pm{\sqrt {2}\over 2},\pm{\sqrt {3}\over 2},{\pm 1\pm\sqrt {5}\over 4}\}$, which contradicts $ t = { - 1\pm\sqrt {17}\over 4}$.

Let $y=2x$, and suppose $2\cos\pi\alpha$ is a root of $f(y)=y^2+y-4=0$. If $\alpha=p/q$ and $\omega=e^{\frac{2\pi i}{2q}}$, however, then because $f$ is irreducible, \[f(y)|g(y)=\prod_{1\le k\le2q,\;(k,2q)=1}(y-\omega^k-\omega^{-k})\in\mathbb{Z}[y].\]But all roots of $g$ have magnitude at most $2$, while $f$ has the root $\frac{-1-\sqrt{17}}{2}$ of magnitude larger than $2$, a contradiction.